Collatz descent for $n \equiv 7 \pmod{32}$

Let $C$ denote the Collatz step map. The assertion is that every $n \equiv 7 \pmod{32}$ has a finite stopping time:

This class is open. Writing $n = 32j + 7$, the first nine steps are parity-determined and carry $n$ to $81j + 20$, which exceeds $n$; the multiplier accumulated so far is $3^{4}/2^{5} = 81/32 > 1$, so no descent has occurred and the parity of the next iterate depends on $j$. Refining the modulus splits the class further without terminating.

The statement is one of four open classes modulo $32$ left by the residue refinement of the Collatz descent problem; the other three are $15 \bmod 16$ and $27 \bmod 32$. Together they carry the whole remaining difficulty of the descent principle, and hence of the Collatz conjecture.

Formalization Note. No positivity hypothesis is needed: $n \equiv 7 \pmod{32}$ forces $n \ge 7$. The index $m$ is unconstrained, but $m = 0$ can never witness the conclusion since $n < n$ is false.